Kinematics

Ø, unless some unexplained special jargon is being used here. [I think, or just guess.]

Nope, no unexplained jargon has been introduced on my part, i copied the question text to text.

Your answer seems to comply with half the class, but at the same time, this professor is an expert in this field, or so he claims. I suppose I should have mentioned that you can consider redundant mechanism as well.

But if the original text has unexplained special jargon, and you simply copy it, it is still unexplained, is it not?

I suppose you could daisy-chain 7 popsicle sticks together, with one end anchored and a mini-motor at each of the six joints, and claim six degrees of freedom. But the whole thing is in one plane, which has only two dimensions. Terminology matters.

Maybe this guy is a kinematic genius, but if the entire class couldn't solve the problem, maybe he's not a pedagogical genius. Sorry.

Out of curiosity, what is the alleged answer?

That is what I am trying to find out and this is why I am in this forum.

A plane actually has 3-DOF, 2 for position, and one for orientation. .

A search on "robotics degrees of freedom" got lots of hits. I haven't explored them yet....

Something still rings weird here. The whole class strikes out, and yet the prof doesn't explain. That would be frustrating, and perhaps hard to justify.

Let us assume a beam and look at one end of it. If the beam is free then there are -in a plane - for the one end 3 degrees of freedom (it can rotate or move in the 2 directions) and for the second with respect to the first less, since the beam has constant length and limits the freedom of second end. Second end can in fact only move on a circle around first considered end. If first end cannot have translations then the beam has only 2 degrees of freedom = 2 x 1 rotation with one of ends as centre, 2 ends thus 2 degrees of freedom.

In the OP were mentioned only rotating possibilities. So that think along above guide line.

I respect the basic principle of CR4 not to give the solution but hep the student find it .

@del the cat.. I think you are a static-titian.. A beam only has one degree of freedom, I would know because I am a kinematian.. :)

It entirely depends on how you define degrees of freedom. (how free is free? What range of movement is required 10mm 10miles?)
You have the x,y,z axes. Do you then add 2 rotational degrees of freedom?
It also depends what you are manipulating and how free is free?
Just look at your fist, it has a good degree of rotation freedom but not 100%.
A single ball joint will give you 2 degrees of rotational freedom from one joint.
The human wrist arguably doesnt use a conventional revolute joint yet give about 180 degrees of rotation.
It's a right pain to discuss this sort of thing without fancy time consuming drawings.
Del

There are already definitions for "degree of freedom " in kinematics:

- freedom has nothing to do with stroke or distance

- depending the space one considers there are different numbers of degrees of freedom: 1-on a line ( a one dimensional space) a point can only move in two direction but because it remains on the line it has only one degree of freedom 2- on a plane ( two dimensional space) a point or any other element (consider an ant) can have 2 translations and one rotation around an axis perpendicular to the plane 3- in a 3 dimensional space an object can have without constrains 3 translations and 3 rotations.

- the number of available degrees of freedom depends on the constrains imposed upon the object. In the OP the constrain was "only revolution ...", which limits the available number of freedom's degrees.

In every "speciality" there are rules and definitions which cannot be changed or elected as one wish. It is a difference between the general study of kinematics and how connections exist in real life where constrains limit the "freedom".

Then what is the answer, smart ass! ;)

To futher add to what nick name has said:

In Kinematics the degrees of freedom in a mechanism can be computed by:

F = 3*(N-1) - 2*L - H

where

F = total degrees of freedom

N = number of links including the frame

L = number of lower pairs (one degree of freedom)

H = number of higher pairs (two degree of freedom).

Here is a link which has several examples (kinematics).

Here is the mechanism, with apologies for my lacking artistic skills.

In my sketch, I've show a simple 'cross' type of mechanism, which has masses on each arm that can slide and rotate, limited only by the intersection. (the 5dof has the whole arrangement being able to rotate as a unit as well, while that option would be grounded in the 4dof version.) In the 4dof version, the center of the cross is fixed (grounded)

Traditionally, you have 6 physical degrees of freedom. 3 in translation, and 3 in rotation. They are quite common in aircraft training, and there is no ambiguity. Traditionally, it refers to a single mass or body such as an aircraft, and not an articulated mechanism. (but since you stipulated 'mechanism', I'll allow it, even though I consider it borderline. I think mine is better.)

Did the original problem allow sliding?

Yes.

Sliding is Translation.

If you apply the equation (which I am sure your prof gave to you) your sketched mechanism has F= 3*(8-1)-2*9-0=3.

This is also from the sketch to see: the final position of the central triangle depends on the 3 input at the motors acting on the 3 beams.

As answer to Tornado's question the mechanism should have ONLY rotating links at least according to the way I understood the initial text. Proposed solution does not correspond to the OP.

That 'solution' is pants, I don't believe it has 4 DOF, as it certainly is not full rotary movement.
Dig a hole and drop a stick in it, that has more DOF and probably only slightly less play.
Del
<click.. sarcasm mode off...
well, it doesn't go right off...
down to minimum....
nah, to hell with it I'll leave it up full>

There are only 6 fundamental degrees of freedom and on a plane,you can have only 2 degrees of freedom viz. translation (prismatic) along one or other of two axes and combination of the two axes. The moment you consider revolute motion about any of the two axes, you are no longer on the given plane constrained as you are because of the other axis. You cannot have revolute joint without involving the third axis (perpendicular to the two axes) on the plane. Your Professor has given you an untenable and insoluble problem.By the same token, you cannot have 5-D O F either.

i do believe you are right .. seems a contradiction .. 5DOF on a planar mechanism ?

You can have a revolute joint involving that third axis. You cannot have motion (translation or rotation) in the third Dimension. (not axis)

Therefore, my diagram is allowable. The OP diagram is incorrect as it uses the same 2 axes 3 times.

Ahh i see .. you could be right too (remember Topol in Fiddler on the roof? i feel like him now) .. see you guys, get me outta here

Please look up #52.Dimensions and axis (Cartesian) are the same giving us a definition of space. The 4^th dimension generally understood is TIME. I do not subscribe to SPACETIME . Einstein has been proved wrong. Incidentally, there is no such thing as TIME, there is only DURATION- a measure of diurnal scale- occuring between two events.

If your ME Prof assigned this project to the class and he still haven't supplied the answer, this is still an homework assignment and still active.

I'll give some direction though. Go to the library and look for "Mechanism and Mechanical Devices Sourcebook" by Nicholas P. Chironis and Neil Slater published by McGraw-Hill

p911

Thanks for your interest in kinematics, the most boring subfield in Mech Eng..

Not really, its fundamentals and can be applied in a number of ways. Stimulates the side of your brain you don't use much.

How does the sketch in post 13 allow more than 3 degrees of freedom?

The triangle in the center is rigid; as a unit it can translate X-Y on the plane (within various limits of travel). It can also rotate about ± 60°. For each such translation/rotation, the positions of all the outer links are predetermined.

If kinematics uses some other concept, it might be zany, but then not boring.

It is a problem of definition:

one degree of freedom means that if you give an input at the specified entry of a mechanical system (machine) the outlet will have a defined movement only function of the input. Several degrees of freedom means that the output is defined by several inputs and cannot be defined by only one!

In general machines have 1 degree of freedom since the designer wants to control the output coherently to the input.

I would suggest to interested participants to read (even down load) the provided link since it is extremely clear as explanation and accessible to every body. This would allow in many situations we discussed a better level of communication. Do not limit at the "kinematic chapter" the others are also very helpful.

Here is the wiki on the subject, which would allow the OP's solution, as it states, withrespect to linkages: "a joint may provide one DOF (hinge/sliding), or two (cylindrical)"

The OP is " Draw a 4-Degree of freedom planar mechanism that has only revolute joints and all the actuators/motors are connected to ground. Draw another one with 5-DOF?"

Your solution is correct from the general point of view but does not respect the OP. You also used in fact elements with 2 degrees of freedom ( rotation + translation).

I think the solution was initiated by the author of the OP but he did not elaborate enough and go as far as he should. His sytem has 3 DOF, now how can one go to 4 and 5? I am sure he can if he came so far, let him have the pleasure to find on his own and start to appreciate kinematics.

You are correct, my solution only has 3-DOF. I have been thinking about it and perhaps i should think along these lines,

if the term degrees of freedom is used to describe the number of parameters needed to specify the spatial pose of a linkage, then i need not necessirly add an x or y rotation to my previous solution.. Is this the way to go..

you said "I think the solution was initiated by the author of the OP but he did not elaborate enough and go as far as he should." .. but which way do i go??

okay.

I think nick-name knows the answer but is holding out for some reason.. i propose another solution, much similar to the one i had before.. it can only translate in x, y and rotate in z, but there are two ways to do rotation.. I am by no means sure if this is correct, but thats what I can think of..

I dunno. All I can think about is using springs in the segments of the central box. (not 'joints' so it should be allowed.) I do better with reality oriented problems.

Chris

That's at least 4: X-Y translation, rotation, and altering the present shape of the central rectangle (which only one more, I think). However, this opens up some ways to generalize further....

This is the reason:

"I am sure he can if he came so far, let him have the pleasure to find on his own and start to appreciate kinematics."

And for 5?

It is a problem of definition:

Funny I said that waaay back...
Del

You are right and you were right. What happened with this thread is what often happens when one tries to bring an answer to a question which can only be solved with a "specific" knowledge and cannot be solved correctly with the general knowledge. In following picture you will find several planar systems with different DOF. This will even better explain what is a "DOF" from the kinematics point of view:

If there are any questions I shall be glad to answer.

This was the reason suggested to read the texts under the recommended link (not by me) which gives a clear view of the different aspects and also shows how "not boring" kinematics can be!

The text in the drawing was too small to read, and enlarging it turned everything into blurry pixels. Reading down the left and then down the right, are the DOFs 1, 2, 3, 0, 1, 3, 2?

Thanks! It's now very legible and clear.

Thank you for sharing your knowledge.

I know that if I ever understood Kinematics like you do, I wouldnt find it to be as boring..

Hi Economist,

I am glad I helped you find a better understanding of this chapter which although not pleasant is quite important in the conception and understanding of mechanical structures.

It is a real pleasure to notice a reaction as yours: it is more worth than any number of GA!

Good luck further and do not hesitate to ask when problems arose you will always find a helping hand.

I agree. I only see 3 DOF represented in it. X-trans, Y-trans, and Z rot.

Chris

GA. Some might want to consider this as 3 revolute joints. It isn't, but it is a good exercise for the reader to figure out why not.

Post 36 gives a potential hint for adding more DOF. This subject can be hard to analyze, as in my oversight in post 3 of rotation in the plane.

Of course, you can put a rotary actuator on any or every joint of these diagrams, but if its position is determined by the other actuators, then it does not add a new DOF. Instead, you just get some actuators fighting with each other.

I would bet that this has occasionally happened in the field of robotics.

To help you think better. If you have a chain of linear elements connected with revolution joints - which ever limit of angle - the position of the chain end if function of the relative position of all beams in the chain. If you have a chain of 1 then the movement of the end is function of only 1 actuator at the basis of the beam. If your chain has 2 beams in series then the end position depends on the angular position of the basis joint AND of the position of the joint BETWEEN the 2 beams, thus it requires 2 defined parameters which means that in this case you have 2 degrees of freedom you should master to obtain a defined position of the chain end.

In following picture you will see how the problem of DOF is:

In robotics there was not fight between motors/actuators only the soft was not written the right way. Such a problem appeared when the 6 degrees of freedom tables for simulators were first developed.

As you see in the case of what the author of the OP proposed the DOF is 3 which ever the Z takes as value.

I wanted to help but I consider that the discussion derived too much from what it should have been so that here is ONE of possible solutions if I did not make an error which is always possible.

Wish I could help but I am having difficulty to understand the question.

You cannot have 4 DOF in a planar mechanism. You can have only 2 DOF i.e., Lin X and Lin Y. ROT X and/or Rot Y will make at least one of the two links go off the XY plane,i.e. into 3 D space, making it NON-PLANAR.

Hi ,

I read with great attention what you wrote but I am afraid you do not read with great attention what other participants write.

It would be positive for your own evolution to try to understand also new aspects of engineering and not stay stuck with aspects which although right are not applicable in this particular case. What you mention is the movement of a point! Here we deal with a number of elements connected by rotative links to each other.

I hope you will agree those are 2 TOTALLY different problems or don't you agree?

I would suggest 1st you read ALL comments before and 2nd you download the link suggested by one of the participants and which VERY good to understand kinematics and not stay out of it.

Original post:

"Draw a 4-Degree of freedom planar mechanism that has only revolute joints and all the actuators/motors are connected to ground. Draw another one with 5-DOF?"

Just imagine two flat bars with longitudinal slots and connected by a vertical pin as a simple assembly. These two bars can slide along each others lengths linearly giving rise to planar movement only, i.e. utilizing only two linear degrees of freedom (x,Y). although, the two bars can rotate about the common pin, the pin represents the Z axis, outside the plane, therefore a NON-PLANAR DOF!! The problem set by the Professor suffers from poor definition!!

Any real solid object cannot remain within the "plane" (comprising only x & Y or x & Z or Y & Z) while rotating about the two axes comprising the plane.

Thanks for your suggestions.